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GMAT Club World Cup 2026 (Day 3): If n is an integer and 3000 > 1/3^n

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RS2302

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08 Jul 2026, 15:58

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I have been given

3000>1/3^n>30

1/3^n can be written as 3^-n

Hence, 3000>3^-n>30

Now, we will look for powers of 3 that satisfy this inequality
Option D

30 lies between 27 and 81 , that means we can say that 3^-n >_3^4.

Similarly, 3000 lies close to 3^7 I.e. 2187

So, we can say that
3^7>_3^-n>_3^4

From this we get -n= 7, 6,5,4 or n= -7,-6,-5,-4

To calculate median where the number of values are even, we take the average of the two middle values

So, -5+-6/2=-11/2 and that’s the answer

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08 Jul 2026, 16:21

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3000 > (1/3)^n > 30

can be rewritten as -

3000 > 3^-n => here only values 7 and below will be right to apply, since 3^8 is 6561 which is more than 3000
and 3^-n > 30 => here only 4 and above will be the right values to apply, since 3^3 is 27 which is not more than 30

Hence range of n is x is -4, -5, -6, -7

median = -5 + (-6) / 2 = -11/2

Answer is D

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08 Jul 2026, 16:31

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find all values of n where 3^-n value is more than 30 but below 3000, they are 4, 5, 6, 7. median is -11/2 since n has negative sign

Bunuel

If n is an integer and 3000 > 1/3^n > 30, what is the median of all possible values of n?

A. -7
B. -13/2
C. -6
D. -11/2
E. -5

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08 Jul 2026, 16:46

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3000 > 1/3^n>30
We know that n is surely negative for 1/3^n to be greater than 30 and n>3 since 30 is greater than 3^3 = 27.
Dividing by 10 throughout
1000> 3^n-1> 10
10^3>3^n-1>10^1
We cab replace 10 with 3^2
3^6> 3^n-1>3^2
Thus, 6>n-1>2
7>n>3
n = -4, -5, -6
Median = -5 (Ans)

Bunuel

If n is an integer and 3000 > 1/3^n > 30, what is the median of all possible values of n?

A. -7
B. -13/2
C. -6
D. -11/2
E. -5

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08 Jul 2026, 17:46

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I think it should have a cleverer solution than mine 🥲.

I approached it by deducing the factor of 3.

3000 > -1/3^n > 30
3*1000 > 3^-n > 3*10
1000 > 3^(-n-1) > 10

Then, I brute-forced the min and max:

Max: 3 must yield a power of 4; 3^2 = 9 is below the lower bound. So, the maximum of n is -n-1 = 3 -> n = -4
Min: For the minimum, things got a bit harder.
- I have the number running in my head: 3^5 = 243, which makes 3^6 around 720, and 3^7 around 1500.
- Which means the power of the minimum bound should stay at 6
- Then, -n-1 = 6 -> n = -7

Now finding the median: (-7) + (-4) / 2 = -11/2

Choice D

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08 Jul 2026, 17:50

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Given 3000 > 3^-n > 30

n should be -ve so that 3 to the power -n is a number > 1
If n=-1, we get 3, but we need a number > 30
Try other values
n=-2, we get 9, not there yet
n=-3, gives 27, still no
n=-4, gives 81 -> valid first value
n=-6, gives 81 * 9 = 729, good
n=-7 gives 729*3 = 2187 good
n=-8 gives 6561 - out of range

So valid values are -4, -5, -6, -7
Median will be (-6-5)/2 = -11/2 D)

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From the equation we can get 2 three things

1. n<0 i.e negative
2. since 1/3^n > 30 => n < -3
3. since 1/3^n < 300 => n >= -7

Leaves 4 values of n => -4, -5, -6, -7 => Median is (-5 - 6)/2 => -11/2

Ans -D

Bunuel

If n is an integer and 3000 > 1/3^n > 30, what is the median of all possible values of n?

A. -7
B. -13/2
C. -6
D. -11/2
E. -5

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08 Jul 2026, 19:33

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30 < 3^-n < 3000

For n = -4, 3^4 = 81.

3^2 approx 10, so 1000 approx 3^6
So, 3000 is approximately 3^7 (rather 3^7 < 3000 because of underestimation)
n is valid for -4, -5, -6, and -7
median = -11/2
Ans D

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It lies between the range of [-4,-6], so median would be -5.

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Given - 3000 > (1/3)^n > 30
Firstly, it is 1/3 so we need n to have negative powers in order to make the resultant number positive. We know 3^3 is 27 so the next number would be higher than 30 so the range would start with "-4". Similarly, to arrive at a number less than 3000 look for powers of 3 that give you a highest number lesser than 3000. It would be 3^7 = 2,187. So n can be minimum "-7" and maximum "-4" or in other words, minimum 81 and max 2,187.
So the range is -7, -6, -5, -4. Since, it is an even numbered range, the median would be the average of the middle two numbers = (-6 + -5) / 2. = -11/2
Answer: D

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So our equation is: 30< (1/3)^n < 3000
Now, 3^3 = 27 => n<= -4
9^3 = 3^6 = 729
3^7 = 2187
3^8 = 6561
=> possible values of n = -4, -5, -6, -7
Median = -5-6/2= -11/2
ANSWER D

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We know n is an integer and reframing the inequality as 30<1/3^n<3000, the values of powers of 3 which keeps us within the range is what matters.
30 lies between 3^3 and 3^4, and 3000 lies between 3^7 and 3^8.

So -n can take values from 4 to 7 (3 or 8 would put us outside the range).
Hence, values of n would be {-7,-6,-5,-4}.
This makes the median -5.5 or -11/2.

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30<3^(-n)<3000....eq1

3^3 is 27
3^4 is 81
3^5 is 243
3^6 is 729
3^7 is 2187
3^8 is 6561

Hence, valid values for supporting eq 1 is -4,-5,-6,-7
Median = (-5-6)/2 = -11/2

Bunuel

If n is an integer and 3000 > 1/3^n > 30, what is the median of all possible values of n?

A. -7
B. -13/2
C. -6
D. -11/2
E. -5

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08 Jul 2026, 22:03

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3^-4
3^-5
3^-6
3^-7

Satisfy the given inequality
Median is (-5-6)/2 == 5.5

Option D

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Bunuel

If n is an integer and 3000 > 1/3^n > 30, what is the median of all possible values of n?

A. -7
B. -13/2
C. -6
D. -11/2
E. -5

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My answer is D.

3 x 10^3 > 3^-n > 3x10
Notices that 3^4 = 81. So n = -4 will be the first value where 3^-n falls in the range between 30 and 3000
Keeps increasing the exponent until n.= -8 where the value will be >6000 and falls outside the range.
So the values of n are -4, -5, -6, -7. Median of this is (-5-6)/2 = -11/2

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Bunuel

If n is an integer and 3000 > 1/3^n > 30, what is the median of all possible values of n?

A. -7
B. -13/2
C. -6
D. -11/2
E. -5

This question was provided by GMAT Club
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Concept Used : A^(-n) = 1/(A)^n and 1/(A)^-n = A^n

we need to realize that is 1/3^n > 1 so n should be -ve as +ve n will further decrease the value of 1/3.

Now 1/3^(-n) = 3^n so for 1/3^n > 30, n should be < -3 as 1/3^(-3) = 3^3 = 27=> n<=-4.
similarly for 1/3^n<3000, n should be >= -7
therefore n can take values : -4, -5, -6, -7
median = (-5-6)/2 = -11/2

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08 Jul 2026, 22:55

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Since n is an integer, we can reduce the equation to integral powers of 3. i.e.

2187 >= 1/3^n >= 81

Hence the possible values of n are -7, -6, -5, -4

Therefore, the median = (-6-5)/2 = -11/2 (Option D is the correct answer)

Bunuel

If n is an integer and 3000 > 1/3^n > 30, what is the median of all possible values of n?

A. -7
B. -13/2
C. -6
D. -11/2
E. -5

This question was provided by GMAT Club
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08 Jul 2026, 23:10

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question ask the median value of n
1. how 1/3^n can be between 3000 and 30 that the clue its negavtive value of n as 1/3^-4= 1/(1/3)^4 which is equal to 81
2. now if we see only 4 value that can satisfy the condition 4,5,6 and 7 because as soon as you go above 3^7 the value will cross upper limit of 3000 hence we have to find median of these four
3 we would have got median directly if number of terms would have odd but as there are 4 terms we have to do (5+6)/2 that is 11/2 and with negative sign.

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08 Jul 2026, 23:18

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Given equation 30 < 1/3^n < 3000

simplyfying this equation we get 30 < 3^(-n) < 3000

so we need to find values of n that should be consitent with the given inequality, and also given that n is negative we need to assume values in -ve so that the the -ve sign is removed

For 30< 3^(-n) , smallest power of 3 to 30 is 3, So n should be greater than -3
For 3^(-n)<3000, power of 3 smaller than 3000 will be -7

therefore the value of n will be -4<=n<=-7
i,e, -4,-5,-6,-7
median will be [(-5)+(-6)]/2 ==> -11/2
Option D is correct

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08 Jul 2026, 23:23

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3000>1/3^n>30= 30<3^-n<3000, what values can we put for -n so it lies between 30 and 3000 , 3^3= 27<30 (no), 3^4=81(ok) so n>=-4, and and for the other end , 3^7= 2187 and 3^8= 6561, so taking n<=2187, so the set will contain (4,5,6,7)= -4<=n<=-7, so for median taking -5+(-6)/2= -11/2

Bunuel

If n is an integer and 3000 > 1/3^n > 30, what is the median of all possible values of n?

A. -7
B. -13/2
C. -6
D. -11/2
E. -5

This question was provided by GMAT Club
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